Question: 1 - What will be the unit digit of the squares of the following numbers?

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Basil Fitzgerald
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1 Square And Square Roots Question: 1 - What will be the unit digit of the squares of the following numbers? (i) 81 Answer: 1 Explanation: Since, 1 2 ends up having 1 as the digit at unit s place so 81 2 will have 1 at unit s place. (ii) 272 Asnwer: 4 Explanation: Since, 2 2 = 4, therefore, square of 272 will have 2 at it's unit place. So, will have 4 at unit s place (iii) 799 Answer: 1 Explanation: Since, 9 2 = 81, So, will have 1 at unit s place (iv) 3853 Answer: 9 Explantion: Since 3 2 = 9, so, will have 9 at unit s place. (v) 1234 Answer: 6 Explanation: Since, 4 2 = 16, so, will have 6 at unit s place

2 (vi) Answer: 9. Explanation: Since, 7 2 = 49. Therefore, will have 9 at unit s place (vii) Answer: 4 Explanation: Since, 8 2 = 64. So, will have 4 at unit s place (viii) Answer: 0 Since, 0 2 = 0. So, will have 0 at unit s place (ix) Answer: 6 Explanation: Since, 6 2 = 36. So, will have 6 at unit s place (x) Answer: 5 Explanation:Since, 5 2 = 25, therefore, will have 5 at unit s place Question: 2. The following numbers are obviously not perfect squares. Give reason. (i) 1057 (ii) (iii) 7928 (iv) (v) (vi) (vii) (viii) Answer: (i), (ii), (iii), (iv), (vi) don t have any of the 0, 1, 4, 5, 6, or 9 at unit s place, so they are not be perfect squares. (v), (vii) and (viii) don t have even number of zeroes at the end so they are not perfect squares.

3 Question: 3. The squares of which of the following would be odd numbers? (i) 431 (ii) 2826 (iii) 7779 (iv) Answer: (i) 431 and (iii) 779. Explanation: (i) and (ii) have odd numbers as their square, because an odd number multiplied by another odd number always results in an odd number. Question: 4. Observe the following pattern and find the missing digits = = = = = = = Explanation: Start with 1 followed as many zeroes as there are between the first and the last one, followed by two again followed by as many zeroes and end with 1.

4 Question: 5. Observe the following pattern and supply the missing numbers = = = = = = = Explanation: Start with 1 followed by a zero and go up to as many number as there are number of 1s given, follow the same pattern in reverse order.

5 Question: 6. Using the given pattern, find the missing numbers = = = _ 2 = _ = _ 2 = _ = = = 43 2 Relation among first, second and third number - Third number is the product of first and second number Relation between third and fourth number - Fourth number is 1 more than the third number

6 Question: 7. Without adding, find the sum. (i) Answer:Since, there are 5 consecutive odd numbers, Thus, their sum = 5 2 = 25 (ii) I Answer:Since, there are 10 consecutive odd numbers, Thus, their sum = 10 2 = 100 (iii) Answer:Since, there are 12 consecutive odd numbers, Thus, their sum = 12 2 = 144 Explanation: = 2 2 = = 3 2 = = 4 2 = = 5 2 = 25 In other words this is a way of finding the sum of n odd numbers starting from 1. Therefore, Sum of n odd numbers starting from 1 = n 2 Question: 8. (i) Express 49 as the sum of 7 odd numbers. Since, 49 = 7 2 So, 7 2 can be expressed as follows:

7 (ii) Express 121 as the sum of 11 odd numbers. Since, 121 = 11 2 Therefore, 121 = Question: 9. How many numbers lie between squares of the following numbers? (i) 12 and = = 169 Now, = 25 So, there are 25-1 = 24 numbers lying between 12 2 and 13 2 (ii) 25 and 26 We know that, 25 2 = 625 And, 26 2 = 676 Now, = 51 So, there are 51-1 = 50 numbers lying between 25 2 and 26 2

8 (iii) 99 and 100 We know that, 99 2 = 9801 And, = Now, = 199 So, there are = 198 numbers lying between 99 2 and 100 2

9 Question: 1. Find the square of the following numbers. (i) = 32 x 32 = 1024 But above method can be tough to calculate. It is easier to calculate such values in the following way:. Since, 32 can be written as (30+2) So, 32 2 = (30+2) 2 = (30+2)(30+2) = 30(30+2)+2(30+2) = x x = = 1024 Answer: 1024 (ii) 35 (35) 2 = (30+5) 2 = (30+5)(30+5) = 30(30+5)+5(30+5) = x x = = 1225 Thus, Answer = 1225 (iii) = (80 + 6) 2 = (80 + 6)(80 + 6) = x x = = 7396 Thus, Answer: 7396

10 (iv) = (90+3) 2 = (90 + 3) (90 + 3) = 90 (90 + 3) + 3 (90 + 3) = x x = = 8649 Thus, Answer: 8649 (v) = (70 + 1) 2 = (70 + 1) (70 + 1) = 70 (70 + 1) + 1 (70 + 1) = x x x 1 = = = = 5041 Thus, Answer: 5041 (vi) = (40+6) 2 = (40 + 6) (40 + 6) = 40 (40 + 6) + 6 (40 + 6) = x x = = = = 2116 Thus, Answer: 2116

11 Question: 2. Write a Pythagorean triplet whose one member is: (i) 6 Solution : As we know 2m, m and m 2-1 form a Pythagorean triplet for any number, m > 1. Let us assume 2m = 6 Therefore, m = 3 And, m = = = 10 And, m 2-1 = = 9-1 = 8 Test: = = 100 = 10 2 Hence, the triplet is 6, 8, and 10 Answer (ii) 14 Solution : Let us assume, 2 m = 14, therefore, m = 7 Now, m = = = 50 And, m 2-1 = = 49-1 = 48 Test: = = 2500 = 50 2 Hence, the triplet is 14, 48, and 50 Answer (iii) 16 Let us assume 2 m = 16, then m = 8 Now, m = = = 65 And, m 2-1 = = 64-1 = 63 Test: = = 4225 = 65 2 Hence, the triplet is 16, 63, and 65 Answer

12 (iv) 18 Let us assume 2 m = 18, therefore, m = 9 Now, m = = = 82 And, m 2-1 = = 81-1 = 80 Test: = = 6724 = 82 2 Question: 1. What could be the possible one s digits of the square root of each of the following numbers? (i) 9801 Answer: 1 and 9. Explanation: Since 1 2 and 9 2 give 1 at unit s place, so these are the possible values of unit digit of the square root. (ii) Answer: 4 and 6 Explanation: Since, 4 2 = 16 and 6 2 = 36, hence, 4 and 6 are possible digits (iii) Answer: 1 and 9 (iv) Answer: 5 Explanation: Since, 5 2 = 25, hence 5 is possible.

13 Question: 2. Without doing any calculation, find the numbers which are surely not perfect squares. (i) 153 (ii) 257 (iii) 408 (iv) 441 Answer: (i) 153 (ii) 257 (iii) 408 Explanation: Since, (i), (ii) and (iii) are surely not be perfect square as these numbers end with 3, 7 and 8. A number can be a perfect square if it ends with 0, 1, 4, 5, 6, 9 only Question: 3. Find the square roots of 100 and 169 by the method of repeated subtraction. Answer: Square root of 100 by Repeated subtraction: = = = = = = = = = = 0 We get 0 at 10th step. Thus, 10 is the square root of 100.

14 Square root of 169 by Repeated subtraction: = = = = = = = = = = = = = 0 We get 0 at 13th step. Thus 13 is the square root of 169

15 Question: 4. Find the square roots of the following numbers by the Prime Factorisation Method. (i) 729 Thus, Answer = 27 (ii) 400 Thus, Answer = 20

16 (iii) 1764 Thus, Answer = 42

17 (iv) 4096 Thus, Answer = 64

18 (v) 7744 Thus, Answer = 88 (vi) 9604 Thus, Answer = 98

19 (vii) 5929 Thus, Answer = 77 (viii) 9216 Thus, Answer = 96

20 (ix) 529 Thus, Answer = 23 (x) 8100 Thus, Answer = 90

21 Question: 5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained. (i) 252 By prime factorisation we get, 252 = 2 x 2 x 3 x 3 x 7 Here, 2 and 3 are in pairs but 7 needs a pair. Thus, 7 can become pair after multiplying 252 with 7. So, 252 will become a perfect square when multiplied by 7. Thus, Answer = 7 (ii) 180 By prime factorisation, we get, 180 = 3 x 3 x 2 x 2 x 5 Here, 3 and 2 are in pair but 5 needs a pair to make 180 a perfect square. 180 needs to be multiplied by 5 to become a perfect square. Thus, Answer = 5 (iii) 1008 By prime factorisation of 1008, we get 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7 Here, 2 and 3 are in pair, but 7 needs a pair to make 1008 a perfect square. Thus, 1008 needs to be multiplied by 7 to become a perfect square Hence, Answer = 7

22 (iv) 2028 By prime factorisation of 2028, we get 2028 = 2 x 2 x 3 x 13 x 13 Here, 2 and 13 are in pair, but 3 needs a pair to make 2028 a perfect square. Thus, 2028 needs to be multiplied by 3 to become a perfect square. Hence, Answer = 3 (v) 1458 By prime factorisation of 1458, we get 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3 Here, 3 are in pair, but 2 needs a pair to make 1458 a perfect square. So, 1458 needs to be multiplied by 2 to become a perfect square. Therefore, Answer = 2 (vi) 768 By prime factorisation of 768, we get 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 Here, 2 are in pair, but 3 needs a pair to make 768 a perfect square. So, 768 needs to be multiplied by 3 to become a perfect square. Hence, Answer = 3

23 Question: 6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained. (i) 252 By prime factorisation of 768, we get 252 = 2 x 2 x 3 x 3 x 7 Here, 2 and 3 are in pair, but 7 has no pair, which can be eliminated after dividing 768 by 7. Hence, 252 needs to be divided by 7 to become a perfect square Thus, Answer = 7 (ii) 2925 By prime factorisation of 2925, we get 2925 = 5 x 5 x 3 x 3 x 13 Here, 5 and 3 are in pair, but 7 has no pair, which can be eliminated after dividing 2925 by 13. Thus, 2925 needs to be divided by 13 to become a perfect square Thus, Answer = 13

24 (iii) 396 By prime factorisation of 396, we get 396 = 2 x 2 x 3 x 3 x 11 Here, 2 and 3 are in pair, but 11 needs another 11 to make a pair. Thus, 11 can be eliminated after dividing 396 by 11. Thus, 396 needs to be divided by 11 to become a perfect square Hence, Answer = 11 (iv) 2645 By prime factorisation of 2645, we get 2645 = 5 x 23 x 23 Here, 23 is in pair, but 5 needs a pair. Thus, 5 can be eliminated after dividing 2645 by 5 Hence, 2645 needs to be divided by 5 to become a perfect square. Therefore, Answer = 5

25 (v) 2800 By prime factorisation of 2800, we get 2800 = 2 x 2 x 7 x 10 x 10 Since, only 7 is not in pair, thus, by eliminating 7, from 2800, it can be a perfect square. Hence, 2800 needs to be divided by 7 to become a perfect square. Therefore, Answer = 7 (vi) 1620 By prime factorisation of 1620, we get 1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5 Here, 2 and 3 are in pair, but 5 is not in pair. Thus, by eliminating 5 from 1620, obtained number will be a perfect square. Hence, 1620 needs to be divided by 5 to become a perfect square Thus, Answer = 5

26 Question: 7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. We need to calculate the square root of 2401 to get the solution. By prime factorisation of 2401, we get 2401 = 7 x 7 x 7 x 7 There are 49 students, each contributing 49 rupees Thus, Answer = 49 Question: plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row. To calculate the number of rows and number of plants, we need to find the square root of By prime factorisation of 2025, we get 2025 = 5 x 5 x 3 x 3 x 3 x 3 There are 45 rows with 45 plants in each of them. Thus, Answer = 45

27 Question: 9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10. Let us find LCM of 4, 9 and 10 4 = 2 x 2 9 = 3 x 3 10 = 5 x 2 So, LCM = 2 2 x 3 2 x 5 = 180 Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a perfect square. The Required number = 180 x 5 = 900 Question: 10. Find the smallest square number that is divisible by each of the numbers 8, 15 and = 2 x 2 x 2 15 = 3 x 5 20 = 2 x 2 x 5 So, LCM = 2 x 2 x 2 x 3 x 5 = 120 As 3 and 5 are not in pair in LCM s factor so we need to multiply 120 by 5 and three to make it a perfect square. Required Number = 180 x 3 x 5 = 2700

28 1. Find the square root of each of the following numbers by Division method. (i) 2304 (ii) 4489 (iii) 529

29 (iv) 3249 (v) 1369 (vi) 5776

30 (vii) 7921 (viii) 576 (ix) 1024

31 (x) 3136 (xi) Find the number of digits in the square root of each of the following numbers (without any calculation). (i) 64 (ii) 144 (iii) 4489 (iv) (v) Answer: If there are even number of digits in square then number of digits in If there are odd number of digits in square then number of digits in (i) 1, (ii) 2, (iii) 2, (iv) 3, (v) 3

32 3. Find the square root of the following decimal numbers. (i) 2.56 (ii) 7.29 (iii) 51.84

33 (iv) (v) Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 402 It is clear that if 2 is subtracted then we will get 400, which is a perfect square.

34 (ii) 1989 Here, 84 X 4 = 336 which is less than 389 And, 85 X 5 = 425, which is more than 389 Hence the required difference = = = 1936 is a perfect square. (iii) 3250 Here, 107 X 7 = 749 is less than X 8 = 864 is more than 750 Hence, the required difference = = = 3249 is a perfect square.

35 (iv) 825 Here, 48 X 8 = 384 is less than X 9 = 441 is more than 425 Hence, the required difference = = = 784 is a perfect square. (v) 4000 Here, 123 X 3 = 369 is less than X 4 = 496 is more than 400 Hence, the required difference = = = 3969 is a perfect square.

36 5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 525 Here, 43 X 3 = 129 is more than X 2 = 84 is less than 125 Hence, required addition = = = 529 is a perfect square.

37 (ii) 1750 Here, 161 X 1 = 161 is 11 more than 150 So, = 1761 is a perfect square (iii) 252 Here, 25 X 5 = 125 is less than X 6 = 156 is more than 152 Required difference = = 4 So, = 256 is a perfect square

38 (iv) 1825 Here, 82 X 2 = 164 is less than X 3 = 249 is more than 225 Required difference = = 24 So, = 1849 is a perfect square

39 (v) 6412 Here, we need 161 X 1 = 161 Required difference = = 149 So, = 6561 is a perfect square 6. Find the length of the side of a square whose area is 441 m². Answer: Area of Square = Side²

40 (a) If AB = 6 cm, BC = 8 cm, find AC Answer = AC 2 = AB 2 + BC 2 = = = 100 (b) If AC = 13 cm, BC = 5 cm, find AB Answer: AB 2 = AC 2 - BC 2 = = = A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this. Here, 61X1=61 is less than X2=124 is more than 100 Hence, the required difference= =39 Min. number of plants required= =961

41 9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement. Here, 42 X 2 = 84 is less than X 3 = 129 is more than 100 Hence, the required difference = = 16 So, 16 children will be left out in the arrangement.

Square & Square Roots

Square & Square Roots 1. If a natural number m can be expressed as n², where n is also a natural number, then m is a square number. 2. All square numbers end with, 1, 4, 5, 6 or 9 at unit s place. All